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HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Hawkes Learning Systems College Algebra.

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Presentation on theme: "HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Hawkes Learning Systems College Algebra."— Presentation transcript:

1 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Hawkes Learning Systems College Algebra Section 8.1: Solving Systems by Substitution and Elimination

2 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Objectives o Definition and classification of linear systems of equations. o Solving systems of equations using substitution. o Solving systems of equations using elimination. o Larger systems of equations. o Applications of systems of equations.

3 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Linear Systems of Equations Many problems of a mathematical nature are most naturally described by two or more equations in two or more variables. A collection of linear equations is called a linear system of equations, or simultaneous linear equations. The three graphs below pictorially identify the three possible varieties of systems of two linear equations in two variables.

4 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Linear Systems of Equations Varieties of Linear Systems The chart below illustrates the only possible solutions to any linear system of equations. If a system of linear equations has No Solutions One Solution Infinite Solutions The system is called InconsistentConsistentDependent

5 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Solving Systems by Substitution The solution method of substitution works by solving one equation and substituting the result into the remaining equations, a method that may require several repetitions. This method can be time consuming for large systems where there are many equations and variables, but for smaller systems, this method works well.

6 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Solving Systems by Substitution For example, we’ll demonstrate solving the system by substitution. We’ll show two (of the four possible) ways of solving this system of equations by substitution. Possibility 1: Possibility 2: Choose an equation. Solve for a variable. Insert the solution into the other equation. Solve for the other variable. Solution is (2,3).

7 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Solving Systems by Substitution If we observe the graph of the two linear equations in the system we solved on the previous slide, we can see that the solution we found, (2,3), does indeed satisfy the system.

8 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 1: Solving Systems by Substitution Solve the system by substitution. Choose an equation. (You could have chosen either!) Solve for a variable. We choose y. Insert the solution found for y into the other equation. Solve for the other variable. In this case, x. Thus, the solution is (1,1).

9 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 2: Solving Systems by Substitution Solve the system by substitution. This is a true and trivial statement which implies that there are an infinite number of solutions. In other words, for any x-value, there exists a y-value that satisfies the system.

10 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Solving Systems by Elimination The method of elimination is based on the goal of eliminating one variable by adding two equations together (some texts call this the addition method). The elimination method works because of the following truth: If and then. Proof: We know that. Since, it must be true that. Similarly, so,. Thus, the above statement, “If and then,” is true.

11 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Solving Systems by Elimination For example, we’ll demonstrate solving the following system by the method of elimination. First, we’ll try adding the equations to see if any variables are eliminated. Adding the equations as they are doesn’t help us, so let’s try manipulating one (or both) of the equations by multiplying both sides of that equation by a constant. If we multiply the first equation by ̶ 3 and add the equations again, the x variable will be eliminated. Continued on the next slide… Not helpful!

12 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Solving Systems by Elimination We found Now we can plug in y into either of the equations in the system. Thus, the solution to the system is.

13 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 3: Solving Systems by Elimination Solve the system by the method of elimination. To eliminate a variable, you’ll need to multiply each equation by a unique constant. Let’s eliminate y. To do so, notice that we’ll have to multiply the first equation by 3 and the second equation by 2. Thus, the solution is.

14 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 4: Solving Systems by Elimination Solve the system by the method of elimination. This equation is false, so the solution to this system is (there are no solutions).

15 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Larger Systems of Equations Algebraically, larger systems of equations can be dealt with in a similar manner to the systems we have been looking at. However, graphically, larger systems are very different. The table below highlights the differences between equations in three variables and equations in two variables. An equation in If a solution exists, it is called an The graph contains The graph is 2 variables ordered pair (x,y) 2 axes2 dimensional 3 variables ordered triple (x,y,z) 3 axes3 dimensional

16 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Larger Systems of Equations Below is a graphical representation of the ordered triple,.

17 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 5: Larger Systems of Equations Solve the system Select two equations and eliminate one variable. We’ll eliminate y from equations ( I ) and ( II ). Continued on the next slide…

18 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Select two other equations and eliminate the same variable, y. We chose equations ( I ) and ( III ). We have now found two linear equations in two variables. Now we will solve those equations by the elimination method. Example 5: Larger Systems of Equations(cont.) Continued on the next slide…

19 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 5: Larger Systems of Equations (cont.) We know that z = –1. We’ll back substitute into one of the linear equations in two variables. We chose the second equation found. Back substitute the values found into an original equation; we chose equation ( I ).

20 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 6: Applications of Systems of Equations Joey has $128 in his pocket in $10, $5, and $1 bills. There are 19 bills in all and he has four more $5 bills than $1 bills. How many bills of each kind does Joey have? x = the number of $10 bills y = the number of $5 bills z = the number of $1 bills Come up with a system of equations from the information given. There are 19 bills. The total value is $128. There are four more $5 than $1 bills. Continued on the next slide…

21 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 6: Applications of Systems of Equations (cont.) Now we can solve the system of equations. Continued on the next slide… Note: One of the equations in the system, y = z + 4 is already an equation in two variables, y and z.

22 HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2011 Hawkes Learning Systems. All rights reserved. Example 6: Applications of Systems of Equations (cont.) Thus, Joey has nine $10 bills, seven $5 bills and three $1 bills.


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